Does sequence produced by Newton's method converge

$\min f=2x_1^2+x_2^2-2x_1x_2+2x_1^3+x_1^4$. Given open ball$B(r)=\{(x_1,x_2)^T:x_1^2+x_2^2<r^2\}$, where$r=\frac{3-\sqrt{3}}{6}$, $\nabla^2f$ is positive definite on $B(r)$.For what starting points $\mathbf{x}_0=(a,a)^T\in B(r)$ does theNewton's method converge?

I derive the sequence $\mathbf{x}_{k+1}=\mathbf{x}_k-(\nabla^2 f)^{-1}\nabla f$, where $\nabla f=(4x_1-2x_2+6x_1^2+4x_1^3,2x_2-2x_1)^T$ ,$\nabla^2f=\begin{pmatrix}12x_1^2+12x_1+4&-2\\-2&2 \end{pmatrix},(\nabla^2f)^{-1}=\frac{1}{24x_1^2+24x_1+4}\begin{pmatrix}2&2\\2&12x_1^2+12x_1+4 \end{pmatrix}$.

However it is obvious that this method gets complicated soon. I am stuck, but I can't think of other methods. Can anyone suggest other ways?

Thanks!